Question

Answer 1

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=kb;c=c=kd\)

Ta có: \(\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2k^2+kb.kd}{d^2k^2-kb.kd}=\dfrac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(1\right)\)

\(\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2+bd}{d^2-bd}\)